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Best of Donald Catlin  # Lay Bets Revisited

5 February 2012

One of the interesting and entertaining things about writing these monthly articles is that I occasionally get a letter from an enlightened reader who casts a novel approach to one of the questions I address. Such is the case in a letter I received from one of my readers named David Korsnack. David addressed the problem I wrote about in my December article entitled All Lay Bet Strategies Are Not Equal. This article addressed questions raised by my friend and colleague The Midnight Skulker. Namely, if two players A and B lay bets, A on the 4 and B on the 4 and 10, how much more frequently does A cycle than B? Refer to my December article for the definition of a cycle. It turns out that the Skulker answered his own question and I presented his solution in my December article. David took an entirely different approach and a very clever one at that. Here is his solution.

Let A represent the expected number of rolls for one cycle in player A's game. If either a 4 or a 7 is thrown, then A's cycle is completed in one roll with probability 9/36. If any other number is rolled, then you are in the same situation as before with one more roll added to the tally so your expected number of rolls is A + 1 with probability 27/36. In symbols

A = (9/36)*1 + (27/36)*(A + 1)

This is a simple linear equation in A and can be solved with A = 4.

Similarly, let B represent the expected number of rolls for one cycle in B's game. If a 7 is thrown then the cycle took one roll with probability 6/36. If a 4 or a 10 is thrown, then we are in player A's situation with one more roll added to the tally, namely, the expected number of rolls is A + 1 with probability 6/36. If any other number is thrown, then we are in the same situation B faced when starting his roll except that we have already used one roll. Thus our expected number of rolls in this situation is B + 1 with probability 24/36. In symbols

B = (6/36)*1 + (6/36)*(A + 1) + (24/36)*(B + 1)

Since we already know that A = 4, this is a simple linear equation in B that can easily be solved with B = 5.

Now suppose that each player takes n rolls (think of n as a large number). Then the expected number of cycles for A is n/A and for B it is n/B. The ratio of the expected number of cycles of A to the expected number of cycles of B is n/A divided by n/B, which is just B/A (the n's cancel) and this number is 5/4. So A will have, on average, 25% more cycles than B. Elegant!

I want to thank David for his letter and his interesting solution to the Skulker's question. Best of all, all I had to do was sit on the sidelines and watch two experts go at it. See you next month.

Don Catlin can be reached at 711cat@comcast.net

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Donald Catlin Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers